Problem 38 - Precalculus

Question

Write the given function as a composition of two or more non-identity functions. (There are several correct answers, so check your answer using function composition.) $$Q(x)=\frac{2 x^{3}+1}{x^{3}-1}$$

Step-by-Step Solution

Answer

Let \( g(x) = x^3 \) and \( f(u) = \frac{2u + 1}{u - 1} \), then \( Q(x) = f(g(x)) \).

Step 1: Analyze the Given Function

The function given is \( Q(x) = \frac{2x^3 + 1}{x^3 - 1} \). We need to express this as a composition of two or more non-identity functions \( f \) and \( g \) such that \( Q(x) = f(g(x)) \).

Step 2: Identify Intermediate Functions

Let’s choose an intermediate function \( g(x) = x^3 \). Essentially, we are breaking down \( Q(x) \) into simpler parts by focusing on the common term under the expressions in the numerator and denominator.

Step 3: Define the Outer Function

Now that we have \( g(x) = x^3 \), we can define the outer function \( f(u) = \frac{2u + 1}{u - 1} \), where \( u = g(x) \). This is derived from substituting \( x^3 \) with \( u \) in \( Q(x) \).

Step 4: Express Q(x) as a Composition

Substitute \( g(x) \) into \( f(u) \), where \( u = x^3 \), to express \( Q(x) \) with these functions: \( Q(x) = f(g(x)) = f(x^3) = \frac{2(x^3) + 1}{x^3 - 1} \), which matches the original function.

Step 5: Verify the Composition

Check: Use the composition \( f(g(x)) = f(x^3) = \frac{2(x^3) + 1}{x^3 - 1} \) to ensure it reconstructs the original function. This confirms that our decomposition is correct.

Key Concepts

Precalculus

Precalculus serves as an essential foundation for higher-level mathematics, including calculus. One of the core fascinations of precalculus lies in its focus on functions, which are the building blocks of mathematical models used in calculus, physics, and engineering. In this context, a function is merely a rule that assigns each input exactly one output.

Function composition, where you combine two or more functions to form a new function, is a crucial topic in precalculus. This is represented as \(f(g(x))\), showing that we apply \(g(x)\) first to our input and then use the result as an input to \(f(x)\). By structuring and deconstructing such compositions, students grasp more complex relationships between variables, enhancing their problem-solving skills. Here, our task was to express \(Q(x)\) as a composition to better understand its structure.

Flexible thinking cultivated through function compositions in precalculus is pivotal for transitioning smoothly into calculus concepts like derivatives and integrals, where such functional operations are regularly used.

Rational Functions

Rational functions are a type of mathematical function, defined as the ratio of two polynomials. A typical rational function can be expressed in the form \(R(x) = \frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) is not zero.

In the exercise at hand, \(Q(x) = \frac{2x^3 + 1}{x^3 - 1}\) is a rational function. Recognizing such functions is important in precalculus because they frequently appear in various mathematical contexts and practical problems. Understanding their behavior involves evaluating how both the numerator and denominator contribute to the overall function.

For instance, the zeros of \(Q(x)\) are derived from the numerator, while the points where the function is undefined are determined by the zeros of the denominator. The composition of functions approach here helps in simplifying complex functions and solving them more easily by breaking them into smaller, more manageable parts.

Intermediate Functions

Intermediate functions serve as a bridge in function composition. They simplify the original function into easier-to-handle components. In our task, choosing \(g(x) = x^3\) as the intermediate function was strategic because it simplifies both the numerator \(2x^3 + 1\) and the denominator \(x^3 - 1\).

This allows us to define another outer function \(f(u) = \frac{2u + 1}{u - 1}\), where \(u = g(x)\). By doing this, the original function is decomposed into two simpler functions that can still represent the original rational function when composed together.

Using intermediate functions is advantageous because it provides a systematic way to tackle more complicated expressions and makes the function analysis more straightforward. This method supports students in better visualizing function behavior and understanding the underlying structure, something that is indispensable in both mathematical theory and application.